Math Problem Statement

Solve the electric field strength of the capacitor's dielectric when the polarization P=2μCm2P=2μCm2, the relative permittivity of the dielectric is εr=5.5εr=5.5 and when the permittivity of the vacuum is known ε0=8.854∗10−12Fm

Solution

To solve for the electric field strength EE in the dielectric of a capacitor, given the polarization P=2μC/m2P = 2 \, \mu C/m^2, relative permittivity εr=5.5\varepsilon_r = 5.5, and permittivity of vacuum ε0=8.854×1012F/m\varepsilon_0 = 8.854 \times 10^{-12} \, F/m, we can use the following relation:

P=ε0(εr1)EP = \varepsilon_0 (\varepsilon_r - 1) E

Where:

  • PP is the polarization,
  • ε0\varepsilon_0 is the vacuum permittivity,
  • εr\varepsilon_r is the relative permittivity of the dielectric, and
  • EE is the electric field strength.

We want to solve for EE:

E=Pε0(εr1)E = \frac{P}{\varepsilon_0 (\varepsilon_r - 1)}

Now, let's plug in the values:

  • P=2×106C/m2P = 2 \times 10^{-6} \, C/m^2
  • ε0=8.854×1012F/m\varepsilon_0 = 8.854 \times 10^{-12} \, F/m
  • εr=5.5\varepsilon_r = 5.5

E=2×1068.854×1012×(5.51)E = \frac{2 \times 10^{-6}}{8.854 \times 10^{-12} \times (5.5 - 1)}

E=2×1068.854×1012×4.5E = \frac{2 \times 10^{-6}}{8.854 \times 10^{-12} \times 4.5}

Now let's calculate EE.The electric field strength EE in the dielectric is approximately 50,197V/m50,197 \, \text{V/m}.

Would you like more details or have any questions about this solution?

Here are five related questions for further exploration:

  1. How does the electric field strength change with varying polarization PP?
  2. What effect does increasing the relative permittivity εr\varepsilon_r have on the electric field strength?
  3. How does the electric field strength compare in a vacuum versus in a dielectric?
  4. Can you calculate the electric field strength if the relative permittivity is doubled?
  5. What is the relationship between capacitance and the electric field in a dielectric material?

Tip: The electric field inside a dielectric material is always weaker than in a vacuum, due to the dielectric's ability to polarize and reduce the field strength.

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Math Problem Analysis

Mathematical Concepts

Electromagnetism
Electric Fields
Capacitors
Dielectrics

Formulas

P = ε₀(εr - 1)E
E = P / [ε₀(εr - 1)]

Theorems

Gauss's Law for Electric Fields in Dielectrics

Suitable Grade Level

College-Level Physics